Question

我有一个链接列表，应该保存每个匹配的结果（W或L）和获得/丢失点数。到目前为止一切都很好，但是当头部不存在/空时我会遇到麻烦。我也意识到我对如何实现链表有一个非常糟糕的概述，任何人都有好的和可理解的资源？无论如何这是我的代码：

#include <stdio.h>
#include <stdlib.h>

struct node {
  int point;
  char outcome;
  struct node *next;
};

void add(struct node *data){
    if(data == NULL){
    data = malloc(sizeof(struct node));
    printf("Outcome and points?\n");
    int point;
    char outcome;
    scanf("%c %d",&outcome,&point);
    fgetc(stdin);
    data->point=point;
    data->outcome=outcome;
    data->next=NULL;
    }else{
        struct node *current= data;
        while(current->next != NULL){
            current = current->next;
        }
        current->next = malloc(sizeof(struct node));
        current=current->next;
        printf("Outcome and points?\n");
        int point;
        char outcome;
        scanf("%c %d",&outcome,&point);
        fgetc(stdin);
        current->point=point;
        current->outcome=outcome;
        current->next=NULL;
    }

}

void print(struct node *data){
    struct node *current = data;
    while(current != NULL){
        printf("%c with %3d\n",current->outcome,current->point);
        current = current->next;
    }
}

int main()
{
    struct node *head=NULL;     
    add(head); 
    add(head);
    add(head); 
    print(head);
}

任何帮助将不胜感激：）

Answer 1

执行时：

void add(struct node *data){
    if(data == NULL){
    data = malloc(sizeof(struct node));

head的值在调用函数中没有变化。

建议改变策略。

struct node* add(struct node *head)
{
   if(head == NULL){
      head = malloc(sizeof(struct node));
      printf("Outcome and points?\n");
      int point;
      char outcome;
      scanf("%c %d",&outcome,&point);
      fgetc(stdin);
      head->point=point;
      head->outcome=outcome;
      head->next=NULL;
   }else{
      struct node *current= head;
      while(current->next != NULL){
         current = current->next;
      }
      current->next = malloc(sizeof(struct node));
      current=current->next;
      printf("Outcome and points?\n");
      int point;
      char outcome;
      scanf("%c %d",&outcome,&point);
      fgetc(stdin);
      current->point=point;
      current->outcome=outcome;
      current->next=NULL;
   }
   return head;
}

然后，改变用法：

int main()
{
    struct node *head = add(NULL);     
    add(head);
    add(head); 
    print(head);
}

Answer 2

您可以通过使用锚节点启动列表来简化代码。锚节点是仅用于其next指针的节点。在下面的代码中，对calloc的调用会创建锚节点，并将锚中的所有字段都设置为0。换句话说，使用next == NULL创建节点。

请注意，在打印列表时，for循环首先跳过锚点节点for (list = list->next;...)

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int point;
    char outcome;
    struct node *next;
};

void add( struct node *list )
{
    struct node *data;

    data = malloc(sizeof(struct node));
    data->next = NULL;
    printf("Outcome and points?\n");
    scanf("%c %d",&data->outcome,&data->point);
    fgetc(stdin);

    while (list->next != NULL)
        list = list->next;
    list->next = data;
}

void print( struct node *list )
{
    for (list = list->next; list != NULL; list = list->next)
        printf("%c with %3d\n", list->outcome, list->point);
}

int main()
{
    struct node *head = calloc( 1, sizeof(struct node) );
    add(head);
    add(head);
    add(head); 
    print(head);
}

附注：我已经省略了一些错误检查以保持简单，您应该真正检查来自calloc，malloc和scanf的返回值并处理任何错误。当然，您应该free最后的所有节点。

链接列表，如果头部不存在，如何插入？

2 个答案: